{1 - \tan \alpha \ast \tan \beta }③利用这些公式可将某些不是特殊角的三角函数转化为特殊角的三角函数来求值,如:\tan 105^{{\circ} }=\tan (45^{{\circ} }+ 60^{{\circ} }) = \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \tan 45^{{\circ} }\ast \tan 60...
{1 + \tan \alpha \tan \beta }(1+ \tan \alpha \tan \beta \neq 0)利用这些公式可以将一些不是特殊角的三角函数转化为特殊角的三角函数来求值.如:\tan 105^{{\circ} }= \tan (45^{{\circ} }+ 60^{{\circ} })= \dfrac{\tan 45^{{\circ} } + \tan 60^{{\circ} }}{1 - \...
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beta +{\sin }^{2}\beta =\dfrac{16}{25}②①+②可得2\left(\sin \alpha \cos \beta +\cos \alpha \sin \beta \right)=-1\therefore \sin \left(\alpha +\beta \right)=-\dfrac{1}{2}\left(2\right)①-②可得\cos 2\beta -\cos 2\alpha +2\sin \left(\alpha -\beta \rig...
阅读下面材料:根据两角和与差的正弦公式,有sin(a+B)sin a cos 8+cos a sin B ---①sin(a-B)=sin a cos B-cos asin B ---②由①+② 得sin(a+B)+sin (a-B)=2 sin acos B ---③令a+B=Aa-B= B 有A+B A-B=22代入③得 A+B A-B sin A+sin B=2 sin COS22.(1)利用上述...
若\sin(\alpha-\beta)\cos\alpha-\cos(\alpha-\beta)\sin\alpha=m,且\beta为第三象限角,则\cos\beta的值为( )。A: \sqrt{1-m^2}B: \sqrt{m^2-1}C: -\sqrt{1-m^2}D: -\sqrt{m^2-1} 相关知识点: 试题来源: 解析 C正确率: 65%, 易错项: B 本题主要考查两角和与差公式及...
sin函数表示一个角度的正弦值，即一个直角三角形的对边与斜边的比值。cos函数表示一个角度的余弦值，即一个直角三角形的邻边与斜边的比值。在力的分解中，经常使用这两个函数来描述力的方向和大小。sin和cos函数还可以用来计算分力的方向角。假设分力Fx和Fy的方向与水平方向的夹角分别为alpha和beta，...
若\( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \),求\( \beta \)的值。 A. \( \frac{\pi}{2} \) B. \( \pi \) C. \( \frac{3\pi}{2} \) D. \( 2\pi \) 相关知识点:
If sin alpha+sin beta=a and cos alpha+cos beta=b show that sin(alpha+beta)=(2ab)/(alpha^(2)+beta^(2)) View Solution Ifcos(θ−α)=aandcos(θ−β)=bthen the value ofsin2(α−β)+2abcos(α−β) View Solution
( \alpha + \beta ) + \sin ( \alpha - \beta )\$ (1);例3求证: \$2 \sin \alpha \cos \beta = \sin ( \alpha + \beta ) + \sin ( \alpha - \beta )\$ (1);例3求证: \$2 \sin \alpha \cos \beta = \sin ( \alpha + \beta ) + \sin ( \alpha - \beta )\...